Items tagged with laplace-transform


My question is: Use the laplace transform to solve the system.

dx/dt + d^2y/dt^2 = 5e^(2t)

dx/dt - x - dy/dt + y = 8e^(2t)

x(0) = 2, y(0) = 1, y'(0) = 1

What I've done in Maple:

eq5 := (diff(x(t), t)+diff(y(t), t$2) = 5*exp(2*t), t, s);

eq5s := laplace(%, t, s);

eq6 := (diff(x(t), t)-x-(diff(y(t), t))+y = 8*exp(2*t), t, s);

eq6s := laplace(%, t, s);

solve({eq5s, eq6s}, {laplace(x(t), t, s), laplace(y(t), t, s)});

subs({x(0) = 2, y(0) = 1, (D(y))(0) = 1}, %);

eq3 := invlaplace(%, s, t);

How do I simplify?  If you plug it into maple I come up with an answer that has x and y on each side.  I guess I'm just wondering how I can set them equal to each other to solve and get rid of the variable x and y.  I know answer is correct as I've also ran it through ODEtest.  Please help.

However you figure out getting rid of the variables I assume will help me also in solving the next problem:

Use the Laplace Transform to solve the system

dx/dt = 7x - y + 6z

dy/dt = -10x + 4y - 12z

dz/dt = -2x + y - z

x(0) = 5, y(0) = 7, z(0) = 2

I have attempted the second problem much like the first.  Thank you for your time.

Is Symbolic Laplace and Inverse Laplace transform possible on Maple? if Yes, how do I find the inverse laplace of this function 


I resolved the coefficients to a 2nd order diff eq of the form:ay''+by'+cy=f(t)

I have included the .mw file for convenience at the link at the bottom of the page.  I resolved the coefficients in 2 different ways & they do not concur.  The 1st approach used the LaPlace transform & partial fraction decomposition.  The coefficient results are given by equations # 14 & 15.  The 2nd approach used undetermined coefficients where I assumed the particular solution and then applied the initial conditions to resolve the coefficients pertaining to the homogeneous solution which are given in the results listed in equation #23.  Noted in the 1st case the coeff's are A3 & A4 and for the 2nd approach the coeff's are A1 & A2.  I have worked this numerous times & do not understand why they do not concur.  So I thought I should get some fresh eyes on the problem to find where I may have gone wrong.

Any new perspective will be greatly apprecieated.

I had trouble uploading the .mw file so I have included an alternative link to retrieve the file if the code contents is illegible or you cannot dowlad the file drectly from the weblink  Download  You should be able to download from the alternative link below once you paste the link into your browser.  If you cannot & wish for me to provide the file in some other fashion respond with some specific instructions & I will attempt to get the file to you.

Thanks 4 any help you can provide.


I would like to apply inverse Laplace transform to U(x,p), which is defined by

For simplicity with my calculations, I assumed p:=i*beta^2. That is why I have the following equation after applying Laplace transform

(beta=0 is not a pole, that is why I removed the last term in my calculations later. Because there is no contribution) where

Here p and beta are complex values, we can write Re(p)=-2*Re(beta)*Im(beta), Im(p)=(Re(beta))^2-(Im(beta))^2 due to p:=i*beta^2. I numerically compute the roots of h(beta), you can find the numerical values of beta (I assumed digits are 50 due to accuracy )

Finally, I would like to plot U(x,t) with the values t=0.8, lambda=1, L=10, k=1. For checking the figure give t=0 and observe that U(x,0)=0.

I am expecting the plot is more or less like the following figure

PS: I already tried to solve and plot the problem, but I could not find where I make a mistake. I  share the worksheet below. Thank you!

I am relatively inexperienced with Maple and would like to either be pointed in the right direction or shown a similar example to my problem for reference.

My problem is I need to write a programme using the below equations



I then have to use the below values to show my programme works






After proving it works I have to apply it to a mass spring system using realistic parameters. Using visual and analytical ways to show findings.



when x and y are the displacement from the equilibrium. Initially the masses are displaced from their equilibrium posistions and released so that

x(0)=a, y(0)=b, Dy(0)=Dx(0)=0

then repeat with different smaller calculations

Any help would be much appreciated


What´s the error here??????

Solução do Sistema de EDO's por Transformada de Laplace




eq1 := diff(x(t), t)+(2.2*x(t)*y(t)+0.5e-1*x(t)/(5+0.5e-1*t)) = 0;

diff(x(t), t)+2.2*x(t)*y(t)+0.5e-1*x(t)/(5+0.5e-1*t) = 0


eq2 := diff(y(t), t)+(2.2*x(t)*y(t)-(0.5e-1*(0.25e-1-y(t)))/(5+0.5e-1*t)) = 0;

diff(y(t), t)+2.2*x(t)*y(t)-0.5e-1*(0.25e-1-y(t))/(5+0.5e-1*t) = 0


eq3 := diff(z(t), t)-2.2*x(t)*y(t)+0.5e-1*z(t)/(5+0.5e-1*t) = 0;

diff(z(t), t)-2.2*x(t)*y(t)+0.5e-1*z(t)/(5+0.5e-1*t) = 0


EQ := [eq1, eq2, eq3]:

for i to 3 do La[i] := laplace(EQ[i], t, s) end do;

s*laplace(x(t), t, s)-1.*x(0.)+2.200000000*laplace(x(t)*y(t), t, s)+laplace(x(t)/(100.+t)^1., t, s) = 0.


-1.*y(0.)-0.2500000000e-1*(exp(100.*s))^1.*Ei(1., 100.*s)^1.+1.*s^1.*laplace(y(t), t, s)^1.+2.200000000*laplace(x(t)^1.*y(t)^1., t, s)+1.*laplace(y(t)^1./(100.+1.*t)^1., t, s) = 0.


s*laplace(z(t), t, s)-1.*z(0.)-2.200000000*laplace(x(t)*y(t), t, s)+laplace(z(t)/(100.+t)^1., t, s) = 0.


LL := subs({laplace(x(t), t, s) = X, laplace(y(t), t, s) = Y, laplace(z(t), t, s) = Z}, [La[1], La[2], La[3]]);

[s*X-1.*x(0.)+2.200000000*laplace(x(t)*y(t), t, s)+laplace(x(t)/(100.+t)^1., t, s) = 0., -1.*y(0.)-0.2500000000e-1*(exp(100.*s))^1.*Ei(1., 100.*s)^1.+1.*s^1.*Y^1.+2.200000000*laplace(x(t)*y(t), t, s)+1.*laplace(y(t)/(100.+t)^1., t, s) = 0., s*Z-1.*z(0.)-2.200000000*laplace(x(t)*y(t), t, s)+laplace(z(t)/(100.+t)^1., t, s) = 0.]


sol := solve(LL, [X, Y, Z]):


SOLS[X, Y, Z]:

SOLT := map(invlaplace, [X, Y, Z], s, t);

[-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(x(_U1)/(100.+_U1), _U1 = 0. .. t))+x(0), -1.*(int(y(_U1)/(100.+_U1), _U1 = 0. .. t))-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))+y(0)+0.2500000000e-1*ln(1.+0.1000000000e-1*t), 2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(z(_U1)/(100.+_U1), _U1 = 0. .. t))+z(0)]


SOLTT := evalf(subs({x(0) = 0.5e-1, y(0) = 0, z(0) = 0}, SOLT));

[-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(x(_U1)/(100.+_U1), _U1 = 0. .. t))+0.5e-1, -1.*(int(y(_U1)/(100.+_U1), _U1 = 0. .. t))-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))+0.2500000000e-1*ln(1.+0.1000000000e-1*t), 2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(z(_U1)/(100.+_U1), _U1 = 0. .. t))]


xx := evalc(Re(SOLTT[1]));

-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(x(_U1)/(100.+_U1), _U1 = 0. .. t))+0.5e-1


yy := evalc(Re(SOLTT[2]));

0.2500000000e-1*ln(abs(1.+0.1000000000e-1*t))-1.*(int(y(_U1)/(100.+_U1), _U1 = 0. .. t))-2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))


zz := evalc(Re(SOLTT[3]));

2.200000000*(int(x(_U1)*y(_U1), _U1 = 0. .. t))-1.*(int(z(_U1)/(100.+_U1), _U1 = 0. .. t))


plot([xx, yy, zz], t = 0 .. 500, legend = [x, y, z]);

Warning, expecting only range variable t in expression -2.200000000*int(x(_U1)*y(_U1),_U1 = 0. .. t)-1.*int(x(_U1)/(100.+_U1),_U1 = 0. .. t)+.5e-1 to be plotted but found names [_U1, x, y]






Warning, solving for expressions other than names or functions is not recommended.

is the warning I get every time I try to solve for solve(ode1 = ode2, theta/T)

Is there any way I can solve for an expression ( theta/T)?


Furthermore I would like to have an answer on how I get to laplace transforms (ODE) with only "s" in the output, if I type like this:

laplace(J*(diff(theta(t), t, t...

Hello all,



As in title I am really bothered with result in floating number obtained from laplace(expr, t, s).

I have a target to laplace and the laplace() gives me result with floating numbers.

(ex. laplace(diff(diff(f(t),t), t), t, s) = laplace(s^2. * laplace(f(t), t, s) : the period after 2 is annoying)


How can I avoid or do some conversions to kill the periods?


I want to invlaplace the following complex expression that I call PQ.


where C1 C3 C4 eta are constant .

Then I do like this


But I got

Good morning everybody,

I have to make the Inverse Laplace Transformation of a function which is too complicated and the only way I have is to make it from the points of the function (I can get a plot but maple cannot get the formal form of the function because it is just too big). Does a numerical method exist in maple which can make the Inversion only from the points of the function?

Thank you,


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